Problem: Find $ \lim_{x\to 6}h(x)$ for $h(x)=\sqrt{5x+6}$.
Explanation: $h$ is a square-root function. Square-root functions are continuous across their entire domain, and their domain is all real $x$ -values for which the expression within the square-root is non-negative. In other words, for any square-root function $q$ and any input $c$ in the domain of $q$ (except for its endpoint), we know that this equality holds: $\lim_{x\to c}q(x)=q(c)$ [What happens at the endpoint?] The input $x=6$ is within the domain of $h$. Therefore, in order to find $ \lim_{x\to 6}h(x)$, we can simply evaluate $h$ at $x=6$. $\begin{aligned} &\phantom{=}h(x) \\\\ &=\sqrt{5x+6} \\\\ &=\sqrt{5(6)+6} \gray{\text{Substitute }x=6} \\\\ &=\sqrt{36} \\\\ &=6 \end{aligned}$ In conclusion, $ \lim_{x\to 6}h(x)=6$.